LC III/475 battery

[arroz]

Hi!

I'm thinking about adapting my LCs to use CR2032 instead of the prone to leak 3.6V batteries. However, I looked into the schematic and realized there is no diode preventing reverse voltage, only a 1K resistor. I powered on one of the Macs without any battery in the slot, and I measured around 2.3V in the battery slot.

AFAIK providing power to a non-rechargeable battery is not great. So… is it safe to use a CR2032? What happens when the battery is drained, below the 2.3V the Mac itself provides when powered on? Is there a possibility the battery catches fire, or shorts internally?

Regards,

Miguel Arroz

 

[saybur]

Can't speak to the Mac electronics here but if you're concerned, adapting a Not-A-Varta to fit the 1/2AA slot might be a good workaround. It's basically just a MAX40200 inline to protect the battery.

Does seem weird that Apple would supply current back into a non-rechargeable battery. I didn't find a spec in a Tadiran 1/2AA datasheet but someone online noted a max 1uA reverse charge on a CR2450 for comparison.

 

[Phipli]

It isn't just exposed straight to a voltage, it is specifically connected to VBATT on U28. I can't find a datasheet for "LS1176" though. Unless it is actually a bus transceiver.

The voltage being there with no battery doesn't necessary mean a significant current would necessarily flow into the battery. You often get a small voltage with absolutely no power behind it in disconnected circuits.

 

[saybur]

That got me curious, apparently the 1176 refers to 343S0120. See here for context.

 

[arroz]

Hi!

Well that at least clarifies that mystery, i was wondering what was that chip.

Yeah this is probably not a problem. I’m just surprised there’s apparently nothing protecting voltage from going in the battery or against the battery being installed the wrong way. Maybe the chip has an internal diode but I would feel better with an external one. Which I can add easily, and take the hit of the voltage drop. I tested and a 3V battery plus diode are enough to make the 475 happy. Not sure for how long, I wonder what’s the minimal voltage that keeps the CUDA alive.

Regards

Miguel Arroz

 

[Phipli]

Hi!

Well that at least clarifies that mystery, i was wondering what was that chip.

Yeah this is probably not a problem. I’m just surprised there’s apparently nothing protecting voltage from going in the battery or against the battery being installed the wrong way. Maybe the chip has an internal diode but I would feel better with an external one. Which I can add easily, and take the hit of the voltage drop. I tested and a 3V battery plus diode are enough to make the 475 happy. Not sure for how long, I wonder what’s the minimal voltage that keeps the CUDA alive.

Regards

Miguel Arroz

Don't add a diode, the coin cells already have a lower voltage than the stock battery and a diode would drop another 0.7v. The chip  will have a diode in it.

I think you might be over thinking things a little. You're replacing a non-rechargable battery with a non-rechargable battery. You don't need to redesign the circuit.